Answer:
The angle of incidence is equal to the angle of reflection.
Explanation:
angle of incidence (i) = angle of reflection (r)
So if the angle of incidence was 45°, the angle of reflection would also be 45°.
When it travels 3m ,4m and 5m it means 12m is right answer.
<span>7.7 m/s
First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so
180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2
Now determine how long you pushed it. For constant acceleration the equation is
d = 0.5 A T^2
Substitute the known values getting,
2.5 m = 0.5 12 m/s^2 T^2
2.5 m = 6 m/s^2 T^2
Solve for T
2.5 m = 6 m/s^2 T^2
0.41667 s^2 = T^2
0.645497224 s = T
Now to get the velocity, multiply the time by the acceleration, giving
0.645497224 s * 12 m/s^2 = 7.745966692 m/s
After rounding to 2 significant figures, you get 7.7 m/s</span>
Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Answer:
O C. Light energy
Explanation:
it conducts energy in it and is an energy itself.
